Find the values of k for which the following pair of linear equations has infinitely many solutions :
` 2x - 3y = 7, ( k + 1 ) x + ( 1 - 2k) y = ( 5k - 4)`.

Explanation

The given equations are
` 2x - 3y - 7 =0`,
` ( k + 1 ) x + ( 1 - 2k ) y + ( 4 - 5k ) = 0 `
These equations are of the form
` a_ 1 x + b_ 1 y + c_ 1 = 0 and a_ 2 x + b_ 2 y + c_ 2 = 0 `
where ` a_ 1 = 2, b_ 1 = - 3, c_ 1 = - 7 `
and `a_ 2 = ( k + 1 ), b_ 2 = (1 - 2k ), c_ 2 = ( 4- 5k )`
Let the given system of equations have infinitely many solutions.
Then, `(a_ 1 ) /(a_ 2 ) = (b_ 1 ) /( b_ 2 ) = (c _ 1 ) /(c_2)`
`rArr ( 2)/(( k+ 1 )) = (-3)/((1- 2k )) = ( -7)/(( 4- 5k ))`
`rArr (2)/(( k + 1 ) ) = (3)/( ( 2k - 1 )) = ( 7)/(( 5k - 4))`
` rArr ( 2)/(( k + 1 )) = ( 3)/(( 2k - 1 )) = ( 7)/(( 5k - 4))`
`rArr ( 2)/((k + 1 )) = (3)/(( 2k - 1 )) and (3)/((2k -1 )) = ( 7)/((5 k - 4))`
`rArr 4k - 2 = 3k + 3 and 15k - 12 = 14k - 7`
` rArr k = 5 and k = 5`
Hence, ` k = 5`